A juggler keeps 5 balls going with one hand, so that at any instant, 4 balls are in air and one ball in the hand. If each ball rises to a height of 20 m, then the time for which each ball stays in the hand is (g=10 m s -2)

Question

A juggler keeps 5 balls going with one hand, so that at any instant, 4 balls are in air and one ball in the hand. If each ball rises to a height of 20 m, then the time for which each ball stays in the hand is (g=10 m s -2) (A) 1 s (B) 1.5 s (C) 2 s (D) 4 s

Tardigrade Admin · Accepted Answer

Here, n=5, x=20 m , g=10 m s -2 Time for one ball in hand, =(T/n-1)=(2 √(2 x/g)/(n-1)) =(2/(n-1)) √(2 x/g) =(2/(5-1)) √(2 × 20/10)=1 s

Q. A juggler keeps 5 balls going with one hand, so that at any instant, 4 balls are in air and one ball in the hand. If each ball rises to a height of 20 m, then the time for which each ball stays in the hand is $(g = 10 m s^{- 2})$

1 s

1.5 s

2 s

4 s

Here, $n = 5, x = 20 m, g = 10 m s^{- 2}$
Time for one ball in hand,
$= \frac{T}{n - 1} = \frac{2 \frac{2 x}{g}}{( n - 1 )}$
$= \frac{2}{( n - 1 )} \frac{2 x}{g}$
$= \frac{2}{( 5 - 1 )} \frac{2 \times 20}{10} = 1 s$

Q. A juggler keeps 5 balls going with one hand, so that at any instant, 4 balls are in air and one ball in the hand. If each ball rises to a height of 20 m, then the time for which each ball stays in the hand is (g=10ms−2)

1 s

1.5 s

2 s

4 s

Here, n=5,x=20m,g=10ms−2 Time for one ball in hand, =n−1T​=(n−1)2g2x​​​ =(n−1)2​g2x​​ =(5−1)2​102×20​​=1s

Q. A juggler keeps 5 balls going with one hand, so that at any instant, 4 balls are in air and one ball in the hand. If each ball rises to a height of 20 m, then the time for which each ball stays in the hand is $(g = 10 m s^{- 2})$

Here, $n = 5, x = 20 m, g = 10 m s^{- 2}$
Time for one ball in hand,
$= \frac{T}{n - 1} = \frac{2 \frac{2 x}{g}}{( n - 1 )}$
$= \frac{2}{( n - 1 )} \frac{2 x}{g}$
$= \frac{2}{( 5 - 1 )} \frac{2 \times 20}{10} = 1 s$